# Annuity Formula and Loan Formula: Solving for Time

– WELCOME TO A LESSON
ON SOLVING FOR TIME IN THE ANNUITY AND LOAN FORMULAS
USING COMMON LOGARITHMS. THIS LESSON DOES ASSUME
YOU’VE ALREADY WATCHED THE VIDEO ON HOW TO SOLVE EXPONENTIAL
EQUATIONS USING COMMON LOGARITHMS. OFTEN WE ARE INTERESTED IN HOW LONG IT WILL TAKE
TO ACCUMULATE MONEY OR HOW LONG WE NEED
TO EXTEND A LOAN TO BRING PAYMENTS DOWN
TO A REASONABLE LEVEL. THIS REQUIRES SOLVING FOR TIME
IN THE ANNUITY OR LOAN FORMULAS, WHICH WE SEE HERE BELOW. IN BOTH CASES,
CAPITAL N IS TIME IN YEARS. TO SOLVE FOR N,
WE’LL USE LOGARITHMS. MORE SPECIFICALLY, WE’LL BE
USING COMMON LOGARITHMS. IN THIS LESSON WE’LL BE USING
THE ANNUITY FORMULA, AND THEN WE’LL SHOW
HOW TO SET UP A SIMILAR PROBLEM USING THE LOAN FORMULA. LET’S BEGIN BY REVIEWING
WHAT THE VARIABLES REPRESENT IN THE ANNUITY FORMULA. P SUB N IS THE BALANCE
IN THE ACCOUNT AFTER N YEARS. D IS THE REGULAR DEPOSIT
OR DEPOSIT EVERY TIME PERIOD. R IS THE ANNUAL INTEREST RATE
AS A DECIMAL, AND K IS THE NUMBER OF
COMPOUNDING PERIODS IN ONE YEAR. LET’S TAKE A LOOK
AT OUR EXAMPLE. IF YOU INVEST \$200 PER MONTH
IN AN ACCOUNT THAT PAYS 3.5% COMPOUNDED
MONTHLY INTEREST, HOW LONG WILL IT TAKE FOR THE
ACCOUNT TO GROW TO \$12,000? LET’S BEGIN BY LISTING OUT
THE GIVEN INFORMATION. IF YOU’RE INVESTING \$200 PER
MONTH, THAT MEANS THAT D IS 200. THE INTEREST RATE IS 3.5%. AS A DECIMAL THAT WOULD BE
0.035, WHICH IS R. INTEREST IS COMPOUNDED MONTHLY, AND SINCE THERE ARE 12 MONTHS
IN A YEAR, K IS 12. WE WANT TO KNOW HOW LONG
IT WILL TAKE FOR THE ACCOUNT TO GROW TO \$12,000. SO P SUB N WOULD BE 12,000. WE DON’T KNOW THE TIME. SO WE’RE TRYING TO SOLVE FOR N, AND NOW, WE’LL SUB THESE VALUES
INTO OUR EQUATION AND THEN SOLVE FOR N. SO NOTICE THAT D=200, K=12, WHICH OCCURS HERE, HERE
AND HERE. R=0.035,
WHICH IS HERE AND HERE, AND P SUB N=12,000. SO AGAIN, OUR GOAL HERE IS TO
SOLVE THIS FOR N, AND BECAUSE N
IS IN THE EXPONENT, THIS IS WHY WE NEED LOGARITHMS
TO SOLVE THIS EQUATION. NOW, TO BEGIN, NOTICE
HOW WE HAVE THIS FRACTION HERE IN THE DENOMINATOR, AND THIS FRACTION BAR MEANS
DIVISION. SO NOTICE HOW IF WE MULTIPLY
BOTH SIDES OF THE EQUATION BY THIS FRACTION HERE, WE CAN ELIMINATE THIS FRACTION
FROM THE RIGHT SIDE. AGAIN, IN THIS FIRST STEP, WE MULTIPLY BOTH SIDES OF THE
EQUATION BY THIS FRACTION HERE. IT’S HELPFUL IF WE CAN THINK OF
THIS AS BEING OVER ONE, AND THEREFORE,
THESE TWO SIMPLIFY TO ONE, AND THIS FRACTION x 12,000=35. LET’S JUST VERIFY THAT. WE HAVE .035 DIVIDED BY 12
AND THEN x 12,000. SO THAT’S WHERE THE 35
CAME FROM. NOW, FOR THE NEXT STEP, IT SAYS
200 x THIS QUANTITY HERE. SO LET’S GO AHEAD AND DIVIDE
BOTH SIDES BY 200, WHICH MEANS THIS WILL SIMPLIFY
TO ONE. SO WE CAN DROP THESE OUTER
PARENTHESES, AND WE’LL LEAVE THIS
AS A FRACTION, EVEN THOUGH WE COULD SIMPLIFY
IT. SO NOW, WE HAVE 35 DIVIDED
BY 200=THIS QUANTITY HERE. AGAIN, NOTICE HOW THIS IS RAISED
TO THE 12N POWER. SO FOR THE NEXT STEP,
WE WILL ADD ONE TO BOTH SIDES. SO WE’D HAVE 35/200 + 1. AGAIN, WE COULD TRY TO SIMPLIFY
THIS, BUT I’M NOT GOING TO, AND THIS IS EQUAL TO THE
QUANTITY 1 + .035 DIVIDED BY 12 RAISED TO THE POWER OF 12N. THIS IS WHERE THE COMMON LOG
IS GOING TO HELP US SOLVE THIS. LET’S GO AHEAD AND PUT THIS
IN PARENTHESES, TOO. WE’RE GOING TO TAKE
THE COMMON LOG OF BOTH SIDES OF THE EQUATION. BY DOING THIS, WE CAN APPLY THE
POWER PROPERTY OF LOGARITHMS HERE, MEANING WE CAN TAKE
THE EXPONENT OF 12N AND MOVE IT TO THE FRONT. SO WE’D HAVE 12N
x THIS LOGARITHM. SO NOW, WE HAVE LOG OF 35
DIVIDED BY 200 + 1=12N x LOG OF 1 + .035
DIVIDED BY 12, AND NOW,
WE CAN ACTUALLY SOLVE FOR N. WE WANT TO ISOLATE N. WE DON’T WANT THIS 12 HERE
OR THIS LOG, WHICH MEANS WE CAN NOW DIVIDE
BOTH SIDES BY 12 LOG OF THE QUANTITY 1
+ .035 DIVIDED BY 12, AND THIS WILL GIVE US
THE VALUE OF N. NOTICE IF WE SIMPLIFY THE RIGHT
SIDE, 12/12 SIMPLIFIES TO 1, AND THESE TWO LOGS ARE THE SAME. SO THIS ALSO SIMPLIFIES TO ONE. WE COULD HAVE CONVERTED SOME OF
THIS TO DECIMALS EARLIER, BUT THEN IF WE USE
ROUNDED DECIMALS TO PERFORM MORE CALCULATIONS, OUR ANSWER WOULD NOT BE
AS ACCURATE. SO NOW, ON THE RIGHT SIDE
WE JUST HAVE N, AND TO EVALUATE THE LEFT SIDE,
WE’LL GO BACK TO THE CALCULATOR. WE DO WANT TO PUT THE ENTIRE
NUMERATOR AND DENOMINATOR IN ITS OWN SET OF PARENTHESES. SO WE’LL HAVE AN OPEN
PARENTHESIS COMMON LOG, WHICH IS HERE, AND THEN WE HAVE
35 DIVIDED BY 200 + 1. NOW, WE’LL NEED A CLOSE
PARENTHESIS FOR THE LOGARITHM AND ANOTHER CLOSE PARENTHESIS
FOR THE NUMERATOR AND THEN DIVIDED BY OPEN
PARENTHESIS FOR THE DENOMINATOR, 12 COMMON LOG 1 + .035
DIVIDED BY 12, CLOSE PARENTHESIS
FOR THE LOGARITHM, ANOTHER CLOSE PARENTHESIS
FOR THE DENOMINATOR AND ENTER. SO IF WE ROUND TO THE NEAREST
TENTH, WE CAN SAY THIS WOULD TAKE
APPROXIMATELY 4.6 YEARS. SO ONCE AGAIN, THIS MEANS THAT
IF YOU INVEST \$200 PER MONTH IN AN ACCOUNT THAT PAYS 3.5%
COMPOUNDED MONTHLY INTEREST, IN APPROXIMATELY 4.6 YEARS, THE BALANCE WOULD BE
APPROXIMATELY \$12,000. NOW, THIS EXAMPLE
WAS FOR AN ANNUITY, BUT THE PROCESS WOULD BE THE
SAME IF WE USE A LOAN FORMULA, EVEN THOUGH THE EQUATION WOULD
LOOK A LITTLE BIT DIFFERENT. SO BEFORE WE GO, LET’S AT LEAST SET ONE UP
THAT INVOLVES A LOAN FORMULA. FOR EXAMPLE, IF YOU PURCHASE A
\$799 TABLET ON YOUR CREDIT CARD, WHICH HAS AN INTEREST RATE
OF 14% COMPOUNDED MONTHLY, HOW LONG WILL IT TAKE FOR YOU
PAY OFF THE PURCHASE IF YOU MAKE PAYMENTS
OF \$75 A MONTH? SO USING OUR LOAN FORMULA
THIS TIME, NOTICE THAT P SUB 0 OR
THE PRINCIPAL WOULD BE \$799. THE MONTHLY PAYMENT, D, IS \$75. THE INTEREST RATE AS A DECIMAL
WOULD BE 0.14, AND BECAUSE THE INTEREST
IS COMPOUNDED MONTHLY, AND THERE’S 12 MONTHS IN A YEAR,
K IS 12. NOW, WE COULD FOLLOW THE SAME
PROCEDURE THAT WE DID FOR THE ANNUITY FORMULA AND SOLVE THIS FOR N 1. jackoshap4